Suppose we have a vector field $f(x, y) = (6x^2, -\sqrt{y})$ and a curve $C$ that is parameterized by $\alpha(t) = \left( t, 4t^2 \right)$ for $-1 < t < 1$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Solution: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = (6x^2, -\sqrt{y})$ and $\alpha(t) = (t, 4t^2)$. $\begin{aligned} &f(\alpha(t)) = \left( 6t^2, -2t \right) \\ \\ &\alpha'(t) = (1, 8t) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_{-1}^1 \left( 6t^2, -2t \right) \cdot (1, 8t) \, dt$ Let's solve the integral. $\begin{aligned} &\int_{-1}^1 \left( 6t^2, -2t \right) \cdot (1, 8t) \, dt \\ \\ &= \int_{-1}^1 6t^2 - 16t^2 \, dt \\ \\ &= \int_{-1}^1 -10t^2 \, dt \\ \\ &= \left[ \dfrac{-10}{3} t^3 \right]_{-1}^1 \\ \\ &= \dfrac{-20}{3} \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = \dfrac{-20}{3}$.